Step 1: (i) Why actinoid chemistry is harder:
In lanthanoids the 4f electrons lie deep inside, so +3 is the common state.
In actinoids the 5f electrons are held less tightly and reach out further. They can join the bonding, so oxidation states from +3 up to +7 appear.
The elements are radioactive and the heavier ones are made only in tiny amounts. This makes experiments difficult.
Step 2: (ii) Why $E^\circ$ for $M^{2+}/M$ varies irregularly:
$E^\circ$ is linked to the energy of the change $M(s) \rightarrow M^{2+}(aq)$.
This energy has three parts: atomization of the solid, removal of two electrons and hydration of the ion.
Each part follows its own trend along the series, and the trends do not cancel smoothly. So the potentials are irregular.
Step 3: (iii) Name the species:
(I) In acid, chromate changes to the orange dichromate ion, $Cr_2O_7^{2-}$.
(II) Cerium forms the +4 state, as in $Ce^{4+}$, which has no f electron left. Terbium also shows +4, but cerium is the standard answer.
Step 4: (iv) Thermal decomposition of potassium permanganate:
When the purple crystals are heated to about 513 K, three products form: green potassium manganate, brown manganese dioxide and oxygen gas.
\[ 2KMnO_4 \rightarrow K_2MnO_4 + MnO_2 + O_2 \]
Step 5: (v) Write the d configurations:
$Cu^+$ is $d^{10}$, $Sc^{3+}$ is $d^0$, $Fe^{2+}$ is $d^6$, $Ti^{3+}$ is $d^1$ and $Zn^{2+}$ is $d^{10}$.
Colour comes from the jump of an electron between d orbitals. This needs at least one d electron and at least one vacant place in the d subshell.
Only $Fe^{2+}$ and $Ti^{3+}$ satisfy this.
Final Answer:
(i) 5f electrons take part in bonding and the elements are radioactive. (ii) Atomization, ionisation and hydration terms vary irregularly. (iii) $Cr_2O_7^{2-}$ and Ce. (iv) $K_2MnO_4$, $MnO_2$ and $O_2$. (v) $Fe^{2+}$ and $Ti^{3+}$ are coloured.